Ay 122 - Fall 2004 Lecture 6 (given by Tony Travouillon) Stellar atmospheres, classification of stellar spectra (Many slides c/o Phil Armitage)
Formation of spectral lines: 1.excitation Two key questions: In what orbital are electrons most likely to be found? What are the relative numbers of atoms in various stages of ionization? Before collision After collision
Formation of spectral lines: 1.excitation From Maxwell-Boltzmann distribution function (statistical mechanics), we can generate the ratio of electrons in different excitation state (boltzmann equation): N g b b -( E -E )/ kt N a = g a e b a Where g is the degeneracy of each level: 2 g n = 2n
Formation of spectral lines: 2.ionization When the temperature become high enough, ionization becomes important, as it can no longer participate in line emission or absorption. When the gas is ionized, electrons and positive ions will recombine. When rates of ionization and recombination are equal, the gas is in ionization equilibrium. Saha s equation: nen( X r+ 1 ) 2gr 2pmr ktk 3/ 2 -[ E / k ] ( ) i kt = e 2 n( X r ) gr+ 1 h
The spectral class and type of a star is directly related to its surface temperature: O stars are the hottest and M stars are the coolest
Most brown dwarfs are in even cooler spectral classes called L and T Unlike true stars, brown dwarfs are too small to sustain thermonuclear fusion
The Hertzsprung-Russell Diagram
Line broadening
Broadening of spectral lines An individual atom making a transition between energy levels emits one photon with a well-defined energy / frequency. However, profiles of real spectral lines are not infinitely narrow. I n Dn e.g. for an emission line, width of the spectral line Dn could be defined as the full width at half the maximum intensity of the line. frequency n Details of definition don t matter - important to see what causes lines to have finite width.
Two basic mechanisms: 1) Energy levels themselves are not infinitely sharp: emitted photons have a range of frequencies 2) Atoms in the gas are moving relative to the observer: observed photons don t have the same frequency as the emitted photons because of the Doppler effect.
Natural linewidth E Ground state Consider excited state with energy E above the ground state. Electrons in excited state remain there for average time Dt before decaying to ground state. Uncertainty principle: energy difference between states is uncertain by an amount DE given by: DEDt ª h 2p But since E = hn, DE = hdn Dn ª 1 2pDt Broadening due to this effect is called the natural linewidth.
Natural linewidth sets absolute minimum width of spectral lines. However, normally very small - other effects dominate. e.g. for hydrogen n=2 to n=1 transition (Lyman a transition) the lifetime is of the order of 10-9 s. Natural linewidth is ~10 8 Hz. Compare to frequency of transition: Dn n ª10-7 In astrophysical situations, other processes will often give much larger linewidths than this.
Collisional broadening In a dense gas, atoms are colliding frequently. This effectively reduces the lifetime of states further, to a value smaller than the quantum mechanical lifetime. If the frequency of collisions is n col, then expect to get a collisional linewidth of about Dn ~ n col. Frequency of collisions increases with density - expect to see broader lines in high density regions as compared to low density ones. e.g. a main sequence star (small radius) has a higher density at the photosphere than a giant of the same surface temperature. Spectral lines in the main sequence star will be broader than in the giant.
Doppler or thermal broadening Atoms in a gas have random motions that depend upon the temperature. For atoms of mass m, at temperature T, the typical speed is obtained by equating kinetic and thermal energy: 1 2 mv 2 = kt k = Boltzmann s const Number of atoms with given speed or velocity is given by Maxwell s law. Need to distinguish between forms of this law for speed and for any one velocity component: z v v x y x v 2 = v x 2 + v y 2 + v z 2 Distribution of one component of the velocity, say v x, is relevant for thermal broadening - only care about motion along line of sight.
For one component, number of atoms dn within velocity interval dv x is given by: Ê dn(v x ) µexp - mv 2 ˆ x Á dv x Ë 2kT Distribution law for speeds has extra factor of v 2 : Ê dn(v) µv 2 exp - mv 2 ˆ Á dv Ë 2kT Most probable speed: v peak = Average speed: 2kT m v rms = v 2 = 3kT m
Emits at frequency n 0 v x Observed at frequency n Consider atom moving with velocity v x along the line of sight to the observer. Doppler shift formula: n -n 0 n 0 = v x c Combine this with the thermal distribution of velocities: f(n) = 1 È Dn D p exp - n -n 0 Í Î Í Dn D ( ) 2 ( ) 2 f where the Doppler width of the line: Dn D = n 0 c 2kT m n
If the gas also has large-scale (i.e. not microscopic) motions due to turbulence, those add to the width: Dn D = n 0 c Ê Á Ë 2kT m + v 2 turb v turb is a measure of the typical turbulent velocity (note: really need same velocity distribution for this to be strictly valid). ˆ 1 2 Some numbers for hydrogen: Dn D Ê ª 4.3 10-5 T ˆ Á n 0 Ë 10 4 K Dn D c Ê T ˆ ª13Á n 0 Ë 10 4 K 1 2 1 2 km s -1 larger than natural linewidth measured in velocity units, comparable to the sound speed in the gas
Summary: Strength of different spectral lines depends upon the abundance of different elements, and on the excitation / ionization state (described in part by the Boltzmann formula). Width of spectral lines depends upon: Natural linewidth (small) Collisional linewidth (larger at high density) Thermal linewidth (larger at higher temperature) High quality spectrum gives information on composition, temperature and density of the gas. c.f. `Modern Astrophysics section 8.1: more on thermal broadening, Boltzmann law, and Saha equation (version of Boltzmann law for ionization).
Stellar Atmosphere
Flux Consider a small area da, exposed to radiation for a time dt. Energy passing through the area is F.dA.dt, where F is the energy flux (units erg s -1 cm -2 ). r da F(r) Unless the radiation is isotropic (same in all directions), F will depend on orientation of da. Spherically symmetric steady source of luminosity L. Energy conservation: L = 4pr 2 F(r) F(r) = L 4pr 2 Inverse square law.
The equation of radiative transfer How does the intensity of radiation change in the presence of emission and / or absorption? Definition of solid angle and steradian q ds Sphere radius r - area of a patch ds on the surface is: ds = rdq rsinqdf r 2 dw dw is the solid angle subtended by the area ds at the center of the sphere. Unit of solid angle is the steradian. 4p steradians cover whole sphere.
Definition of the specific intensity Construct an area da normal to a light ray, and consider all the rays that pass through da whose directions lie within a small solid angle dw. Solid angle dw da The amount of energy passing through da and into dw in time dt in frequency range dn is: de = I n dadtdndw Specific intensity of the radiation.
Compare with definition of the flux: specific intensity is very similar except it depends upon direction and frequency as well as location. Units of specific intensity are: erg s -1 cm -2 Hz -1 steradian -1 Same as F n Another, more intuitive name for the specific intensity is brightness.
Simple relation between the flux and the specific intensity: Consider a small area da, with light rays passing through it at all angles to the normal to the surface n: n I n If q = 90 o, then light rays in that direction contribute zero net flux through area da. q For rays at angle q, foreshortening reduces the effective area by a factor of cos(q). Hence, net flux in the direction of n is given by integrating (the specific intensity x cos q) over all solid angles: F n = Ú I n cosqdw Note: to actually evaluate this need to express dw in terms of dq and df as before.
How does specific intensity change along a ray If there is no emission or absorption, specific intensity is just constant along the path of a light ray. Consider any two points along a ray, and construct areas da 1 and da 2 normal to the ray at those points. How much energy is carried by those rays that pass through both da 1 and da 2? da 1 da 2 de 1 = I n1 da 1 dtdn 1 dw 1 de 2 = I n 2 da 2 dtdn 2 dw 2 where dw 1 is the solid angle subtended by da 2 at da 1 etc
The same photons pass through both da 1 and da 2, without change in their frequency. Conservation of energy gives: de 1 = de 2 dn 1 = dn 2 - equal energy - same frequency interval Using definition of solid angle, if da 1 is separated from da 2 by distance r: Substitute: dw 1 = da 2 r, dw 2 2 = da 1 r 2 I n1 da 1 dtdn 1 dw 1 = I n 2 da 2 dtdn 2 dw 2 da I n1 da 1 dtdn 2 da 1 = I r 2 n 2 da 2 dtdn 1 2 r 2 I n1 = I n 2 de 1 =de 2 dn 1 =dn 2
Conclude: specific intensity remains the same as radiation propagates through free space. Justifies use of alternative term `brightness - e.g. brightness of the disk of a star remains same no matter the distance - flux goes down but this is compensated by the light coming from a smaller area. If we measure the distance along a ray by variable s, can express result equivalently in differential form: di n ds = 0
Emission If the radiation travels through a medium which is itself emitting radiation, that will add to the energy: I n I n +di n da ds Spontaneous emission coefficient is the amount of energy emitted per unit time, per unit solid angle, per unit frequency interval, and per unit volume: de = j n dvdwdtdn In going a distance ds, beam of cross-section da travels through a volume dv = da x ds.
Change (increase) in specific intensity is therefore: di n = j n ds Equation of radiative transfer for pure emission becomes: di n ds = j n If we know what j n is, can integrate this equation to find the change in specific intensity as radiation propagates through the gas: s Ú I n (s) = I n (s 0 ) + j n ( s )d s s 0 s=s 0 s i.e. add up the contributions to the emission all along the path.
Absorption If the radiation travels through a medium which absorbs (or scatters) radiation, the energy in the beam will be reduced: I n I n +di n da ds Number density of absorbers (particles per unit volume) = n Each absorber has cross-sectional area = s n (units cm 2 ) If beam travels through ds, total area of absorbers is: number of absorbers cross - section = ndads s n
Fraction of radiation absorbed = fraction of area blocked: di n = - ndadss n = -ns n ds I n da di n = -ns n I n ds -a n I n ds absorption coefficient (units cm -1) Can also write this in terms of mass: a n rk n k n is called the mass absorption coefficient or the opacity. Opacity has units of cm 2 g -1 (i.e. the cross section of a gram of gas).
Example: Thomson scattering A free electron has a cross section to radiation given by the Thomson value: s n T = 6.7 10-25 cm 2 independent of frequency. The opacity is therefore: k n = n r s n = N As n = 0.4 cm 2 g -1 If the gas is pure hydrogen (protons and electrons only) (note: really should distinguish between absorption and scattering, but don t need to worry about that here )
Equation of radiative transfer for pure absorption. Rearrange previous equation: di n ds = -a n I n Different from emission because depends on how much radiation we already have. Integrate to find how radiation changes along path: s s di Ú n = - Ú a n ( s )d s s0 I n s 0 [ ln I ] s n s0 s Ú = - a n ( s )d s s=s 0 s s 0 I n (s) = I n (s 0 )e - a n ( s0 s Ú s )d s
e.g. if the absorption coefficient is a constant (example, a uniform density gas of ionized hydrogen): I n (Ds) = I 0 e -a n Ds Specific intensity after distance Ds Initial intensity Radiation exponentially absorbed with distance Radiative transfer equation with both absorption and emission: di n ds = -a n I n + j n absorption emission
Optical depth Look again at general solution for pure absorption: I n (s) = I n (s 0 )e - a n ( s 0 s )d s Imagine radiation traveling into a cloud of absorbing gas, exponential defines a scale over which radiation is attenuated. s Ú When: s Ú s 0 a n ( s )d s =1 s=s 0 intensity will be reduced to 1/e of its original value.
Define optical depth t as: s Ú t n (s) = a n ( s )d s or equivalently s 0 dt n = a n ds A medium is optically thick at a frequency n if the optical depth for a typical path through the medium satisfies: t n 1 Medium is said to be optically thin if instead: t n <1 Interpretation: an optically thin medium is one which a typical photon of frequency n can pass through without being absorbed.
Can rewrite the radiative transfer equation using the optical depth as a measure of `distance rather than s: di n ds = -a n I n + j n di n a n ds = -I n + j n a n divide by the absorption coefficient di n dt n = -I n + S n where S n = j n / a n is the source function. An alternative and sometimes more convenient way to write the equation.
Use result we derived last time - consider radiation passing through a hot cloud of gas in thermal equilibrium: I 0 I n (t) Found: I n (t n ) = I 0 e -t n + B n (1- e-t n ) Suppose no intensity entering the cloud, I 0 = 0. If the cloud is very optically thin: e -t n ª1- t n I n (t n ) = B n (1-1+ t n ) = t n B n
Optical depth is related to the absorption coefficient via: Means that: t n = a n Ds (for constant a) I n = t n B n µa n B n Intensity is large at frequencies where the absorption coefficient is large. For a hot gas, absorption coefficient is large at the frequencies of the spectral lines. For an optically thin medium such as a nebula, expect an emission line spectrum with large intensity at the frequencies where a n is large.
Summary: Friday s class I 0 I n (t) Used solution: I n (t n ) = I 0 e -t n + B n (1- e -t n ) For small optical depth: I n = t n B n µa n B n For an optically thin medium such as a nebula, expect an emission line spectrum with large intensity at the frequencies where a n is large.
Next, consider an optically thick source: Already shown that in the interior, radiation will be described by the Planck function. Radiation escaping from the source will be modified because the temperature (and thus the Planck function) varies along the path. Example: model a star using a two layer model: T in T out Radiation starts from the inner layer as blackbody radiation at temperature T in. Escapes through an atmosphere of optical depth t and temperature T out.
Use same solution as before to describe change in intensity of the radiation: I n (t n ) = I 0 e -t n + B n (1- e -t n ) Escaping radiation B n (T in ) B n (T out ) Valid provided that all the gas is in thermal equilibrium (LTE). Assume that optical depth of outer layer is small and use approximate expansion for the exponential as before: I n (t n ) = B n (T in )e -t n + B n (T out )[ 1- e -t ] n I n (t n ) = B n (T in ) 1- t n [ ] + B n (T out ) t n [ ] I n (t n ) = B n (T in ) + t n B n (T out ) - B n (T in )
I n (t n ) = B n (T in ) + t n [ B n (T out ) - B n (T in )] Initial radiation intensity Change in intensity caused by the outer layer. Depends upon frequency. Recall that intensity of blackbody radiation increases at all frequencies as the temperature goes up. Sign of the second term depends upon whether B n (T out ) is larger or smaller than B n (T in ) - i.e. on whether T out > T in.
I n (t n ) = B n (T in ) + t n [ B n (T out ) - B n (T in )] 1) T out > T in : second term is positive: Escaping intensity is larger at frequencies where t n is greatest (frequencies corresponding to spectral lines). Expect emission lines on top of the continuum. 2) T out < T in : second term is negative: Escaping intensity is reduced at frequencies where t n is greatest (frequencies corresponding to spectral lines). Expect absorption lines superimposed on the continuum.
For the Sun, temperature near the optical photosphere decreases outward (as it must since energy transport is from the center to the outside). In second regime: T out < T in Expect to see an absorption spectrum, as observed:
Note: see strong UV and X-ray emission from the Solar corona, so obviously the temperature there is much hotter than that of the photosphere T UV radiation comes from region where T increasing, so emission line spectrum radius Optical radiation comes from region where T decreasing, so absorption spectrum